Complex formation metal ion=lewis acid=electron-pair acceptor ligand = Lewis base= electron-pair donor

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Myron Tate
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Complex formation titrations Complex formation Chelate effect EDTA Conditional formation constant EDTA titration Indicators Types of titrations Auxiliary complexing agents Complex formation metal

1 Complex formation titrations Complex formation Chelate effect EDTA Conditional formation constant EDTA titration Indicators Types of titrations Auxiliary complexing agents Complex formation metal ion=lewis acid=electron-pair acceptor ligand = Lewis base= electron-pair donor Chelate effect Multidentate ligands form more stable complexes than monodentate ligands In general, Kf is larger for polydentate ligands more complete reaction with cation tend to form 1:1 complex sharper end point EDTA H 6 Y 2+ H 5 Y + + H + pk 1 = 0.0 H 5 Y + H 4 Y + H + pk 2 = 1.5 H 4 Y H 3 Y - + H + pk 3 = 2.0 H 3 Y H 2 Y - + H + pk 4 = 2.66 H 2 Y 3- HY + H + pk 5 = 6.16 HY 3- Y + H + pk 6 = H 4 Y = zwiterrion

2x10 18 If both Ag + + Ni 2+ are present at ~ equimolar concentration: complex formation in the order NiY 2 then AgY 3 Conditional

2 EDTA uncomplexd EDTA complexes M n+ + Y 4- (n-4) MY formation constant or stability constant (Table 11-2) Examples Ag + + Y 4-3- AgY K AgY = 2.1x10 7 Ni 2+ + Y 4 NiY K NiY = 4.2x10 18 If both Ag + + Ni 2+ are present at ~ equimolar concentration: complex formation in the order NiY 2 then AgY 3 Conditional formation constant complex formation is always in competition with protonation of Y 4- ph control is important use of a buffer is mandatory = K ' f = effective or conditional formation constant Constant at fixed ph and fixed ionic strength

3 EDTA titration Titrant H 4 Y (dry at o C several hours) Na 2 H 2 Y.2H 2 O (dry at 80 o C several days) Before equivalence point: excess M n+ At equivalence point: Y 4 + M n+ (n-4)+ MY [M n+ ]=[EDTA]=[Y 4- ]+[HY 3- ]+[H 2 Y ]+[H 3 Y 1- ]+[H 4 Y] eq i alence point After equivalence point: excess EDTA End point detection: Chelating agent less strongly bound to Mn+ than Y4- Ion selective electrode Example: conc. of free Fe 3+ in 0.01 M FeY- at ph 8.00 and 2.00 Fe 3+ + Y 4- - FeY At low ph higher Fe because more competition K FeY = [FeY - ]/[Fe 3+ ][Y 4- ] = 1.3 x for EDTA from hydronium at ph 9.00: K FeY = (5.4x10-3 ) (1.3x10 25 ) = 7.0 x α 4 K MY at ph 2.00: K FeY = (3.7x10-14 )(1.3x10 25 ) = 4.8 x α 4 K MY Example: conc. of free Fe 3+ in 0.01 M FeY - at ph 8.00 and 2.00 Fe 3+ + Y 4- FeY- K FeY = [FeY - ]/[Fe 3+ ][Y 4- ] = 1.3 x at ph 9.00: K FeY = (5.4x10-3 ) (1.3x10 25 ) = 7.0 x α 4 K MY at ph 2.00: K FeY = (3.7x10-14 )(1.3x10 25 ) = 4.8 x α 4 K MY Fe 3+ + Y 4- FeY M initially x x 0.10-x at equilibrium at ph 8 00 K FeY = [FeY - ]/[Fe 3+ ][Y 4- ] = = (0.10-x)/x 2 [Fe 3+ ] = x = 1.2x10-12 M at ph 2.00 [Fe 3+ ] = x = 4.8x10-7 M Metal-EDTA complex less stable at low ph Ensure something is dry: weigh it, put it in a desiccator, and weigh after - if lost weight: not dry then heat it (if stable) and keep going until doesn't lose weight

Effective titration goes to completion large Kf required high ph favours α4 less competition from protonation of EDTA increases K'f possible to titrate selectively a metal ion in presence of another

4 Effective titration goes to completion large Kf required high ph favours α4 less competition from protonation of EDTA increases K'f possible to titrate selectively a metal ion in presence of another one Example: Fe 3+ (log K f =25.1), Ca 2+ (log K f =10.65) titrate at ph 4.0 Fe 3+ only α 4 = 3.0x10-9 K' f = 3.8x10 16 for Fe 3+ (vs 134 Ca 2+ ) Requirement minimum K f ' for 99.9% complete titration: Mn + + Y 4- MY (n-4) F 10-3 F F-10-3 F at equivalence point IF F = 10-2 M K MY = 10 8 Need to have a K F that is 10 6 /F Which of the following EDTA titrations would be most complete? a) Cu 2+ at ph 10 Need to calculate conditional formation constant: b) Al 3+ at ph 4 Cu: K F = 1.8x10 18 x 0.30 = 1.8x10 18 c) Ti 3+ at ph 5 Ti: 2.00x10 21 x 2.9x10-7 = 5.8x10 14 Al: 2.5x x10-9 = 7.5x10 7 α 4 =3.0x10-9 at ph 4.0 log K f for Cu 2+ = α 4 =2.9x10-7 at ph 5.0 log K f for Al 3+ = 16.4 α 4 =0.30 at ph 10 log K f for Ti 3+ = 21.3 EDTA titration curve example: 50.0 ml M Mg 2+ titrated with M EDTA buffered at ph Mg 2+ + Y 4- MgY K MY ' = α 4 K MY = (0.36)(6.2x10 8 ) = 2.2x10 8 large ie. Goes to completion after each addition of EDTA V e =50.0 ml x M/ M = 50.0 ml EDTA Example (continued) Before the equivalence point: after 5.0 ml EDTA added: excess Mg 2+ Fraction left Original conc. Dilution factor pmg 2+ = - log [Mg 2+ ] = 1.39 express at two decimal places by convention

5 at the equivalence point: formal conc: Mg 2+ + Y 4- MgY M initially x x x at equilibrium pmg 2+ = 4.97 After the equivalence point excess of EDTA pmg increasing as do titration [Mg 2+ ] = 2.2x10-7 M pmg 2+ = 6.65 Indicators for EDTA titrations Eriochrome black T (ECBT) H 2 In - + H 2 O HIn + H 3 O + pk a2 = 6.3 red blue HIn + H 2 O In + H 3 O + pk a3 = 11.6 blue orange Mg 2+ + In 3+ MgIn - K f = 1.0x10 7 = [MgIn - ]/ [Mg 2+ ][In 3- ] colourless orange wine-red Calmagite more stable than ECBT pk 2 = 8.1, pk 3 = 12.4 (same colour changes) See Table 11-3 for more indicators Conditions for use of indicator Important to adjust ph such that: MgIn - + EDTA HIn + MgY wine-red colorless blue colorless Also K MIn < K MY otherwise direct titration is not possible K MIn > K MY for Cu 2+, Ni 2+, Co 2+, Cr 3+, Fe 3+, Al 3+ with ECBT During a titration: EDTA reacts with free Mn + first: no color change then with MIn (n-3) color change Want displacement of indicator by metal: Want formation complex for indicator-edta to be smaller than formation complex for metal- EDTA complex

6 A student wishes to use Eriochrome black-t as an indicator for an EDTA titration of Pb 2+, and therefore, must work at a ph of about 10. To adjust the ph, the student should use a) NaOH. b) NH 3. c) NH 3 buffer Direct titration use if analyte + EDTA complex reaction to completion K' f = 10 6 /F for 99.9% completion indicator of K MIn < K MY Back-titrations Required if: analyte precipitates in absence of EDTA analyte reacts too slowly with EDTA analyte blocks the indicator (i e no indicator available) Add known amount of EDTA Back-titrate the excess with metal ion (Mg 2+, Zn 2+...) using a suitable indicator metal ion used for back titration forms weaker EDTA complex than the analyte-edta complex When is a direct EDTA titration not useful? a) When the metal precipitates in the absence of EDTA. b) When the metal reacts too quickly with EDTA. c) When the metal does not block the indicator. Displacement titrations if no satisfactory indicator is available; example: Hg 2+ complexed form add unmeasured excess of a metal-edta complex with Kf < that of analyte-edta complex Example: Mn + + MgY MY n-4 + Mg 2+ 2Ag + + Ni(CN) Ag(CN) 2 + Ni 2+ Titrate displaced Mg2+ with EDTA and ECBT (or other suitable indicator) Which of the following displacement titrations would be feasible? a) Ag + + MgY Mg 2+ + AgY 3- Need formation constant to b) Ba 2+ + MgY Mg 2+ + BaY be greater than magnesium's c) Cu 2+ + MgY 2+ Mg + CuY log K f = 7.20 (Ag + ), 7.88 (Ba 2+ ), (Cu 2+ ), 8.79 (Mg 2+ ) Doesn't matter if excess: doesnt matter how much extra metal-edta complex have present titrating displaced magnesium, extra will stay in Indirect titration anions (CO 3 CrO 4 S SO 4 ) that precipitate with certain metal ions can be determined indirectly with EDTA Add excess metal to precipitate all the anion, filter and wash precipitate. Either boil precipitate with excess EDTA to dissolve the precipitate (form MY ) and then backtitrate excess EDTA with Mg 2+ (less used) Or titrate excess metal ion in the filtrate with EDTA.

7 Masking agents used to increase the selectivity of a titration potentially interfering elements can be masked i.e. complexed with a ligand more stable than EDTA complex free element titrated masked one is demasked and titrated Examples: Al F - AlF6 3+ very stable allows Mg 2+ + Y 4- MgY to take place. CN - complexes Cd 2+, Zn 2+, Hg 2+, Co 2+, Cu +, Ag +, Ni 2+ Pd 2+, Fe 2+, Fe 3+ but NOT Mg 2+, Ca 2+, Mn 2+, Pb 2+ Fe3+ can be successfully titrated with EDTA in the presence of Ni2+ by a) masking Ni 2+ with cyanide and titrating Fe 3+ at ph 3.0. b) titrating at ph 2.0. Conditional formation c) masking Ni 2+ with triethanolamine and titrating Fe 3+ at ph 3.0. constant much larger α 4 = 2.6x10-14 at ph 2.0 log K f for Zn = 25.1 for iron than nickel α 4 = 2.1x10-11 at ph 3.0 log K f for Ni = titrate iron, then K' f = 3.3x10 11 for Fe 3+ >> K' f = 6.5x10 4 for Ni 2+ nickel Auxiliary complexing agents Have to make sure Used to prevent precipitation of analyte as hydrous oxide formation constant Effect on conditional formation constant? α = [Mn + ]/C m C m formal concentration of free metal (ie excluding MY concentration) Example: addition of NH 3 for titration of Zn 2+ Zn 2+ + NH 3 ZnNH ZnNH 3 + NH 3 Zn(NH3) 2 Zn NH 3 Zn(NH3) 2 Beta: addition of two ligands successfully Effect of NH 3 on EDTA titration of Zn Zn NH 3 Zn(NH3) 3 2+ β 3 = [Zn(NH 3 ) 2+ 3]/[Zn 2+ ] [NH 3 ] 3 Zn NH 3 Zn(NH3) 4 β 4 = [Zn(NH 3 ) 4 ]/[Zn 2+ ][NH 3 ] 4 C M = [Zn ] + [ZnNH 3 ] + [Zn(NH 3 ) 2 ] + [Zn(NH 3 ) 3 ] + [Zn(NH 3 ) 4 ] C M = [Zn 2+ ]+ K 1 [NH 3 ][Zn 2+ ]+ β 2 [NH 3 ] 2 [Zn 2+ ] + β 3 [NH 3 ] 3 [Zn 2+ ] + β 4 [NH 3 ] 4 [Zn 2+ ] Only auxilary compound and formation constants Since [Mn + ] = α M C M EDTA complex formation is affected

8 Effect of NH 3 on EDTA titration of Zn (continued) thus conditional constant for a give ph and a given [NH 3 ] C M = total conc. of metal uncomplexed with EDTA C T = total conc. of uncomplexed EDTA Since α M < 1 K'' MY < K' MY less sharp end point keep [NH 3 ] to the minimum needed Requirement: K MY > that of complex between metal and auxiliary complexing agent Example: Titration of ml M Zn 2+ with M EDTA at ph in presence of 0.10 M NH 3 (in that form; NH 4+ also there) V e = ml EDTA α Zn2+ = 1.8x10-5 α 4 = 0.36 K'' MY = α Zn2+ α 4 K MY = ( )(0.36)( )= log K 1 =2.4, log K 2 =2.4, log K 3 =2.5, log K 4 = 2.1: stepwise formation constants for 4Zn 2+ -NH 3 complexes Example (continued) 1) After ml of EDTA: excess Zn 2+ = M [Zn 2+ ]= α Zn2+ C Zn2+ = ( )( M) = M pzn 2+ = -log [Zn 2+ ] = ) At equivalence point: x = C Zn2+ = M [Zn 2+ ] = α Zn2+ C Zn2+ = M pzn = 12.05

9 3) After ml of EDTA = M [ZnY ] = (50.00 ml M)/110mL = M Since both [EDTA] AND [ZnY] are known: pzn = another method: K''Zn = 2.05x10 11 = [ZnY ]/[Zn 2+ ][EDTA] = 4.545x10-4 / x 9.09x10-5 C Zn2+ = 2.439x10-11 [Zn 2+ ] = alpha M Cz n2+ = 1.8x10-5 = 4.4x10-6 M pzn = -log of above = 15,36 both ways give the same answer Example (continued) Past equivalence point [NH 3 ] does not affect pzn but ph does (affects α 4 ) If too much NH 3 is used: no more end point seen Indicator: must have: K ZnIn >K Zn(NH3)x (ECBT OK) so that indicator can displace ammonia K ZnY > K In > K ZnNH3x reacts with amonia first, then indicator, then complexing agent The presence in the solution of an auxiliary complexing agent such as tartrate ion, at a given concentration of Y 4- would a) have no effect on [MY n-4 ]. b) lower [MY n-4 ] c) increase [MY n-4 ].

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